(a) The position vectors of points L and M are (5i + 6j) and (13i + 4j) respectively. If point K lies on LM such that LK : KM is 2 : 3, find the position ve...

Answer Details

(a) K divides LM with LK:KM = 2:3, so the position vector of K is the weighted mean \[ \vec{K} = \frac{3\vec{L} + 2\vec{M}}{5}. \] Thus \[ \vec{K} = \frac{3(5\mathbf{i}+6\mathbf{j}) + 2(13\mathbf{i}+4\mathbf{j})}{5} = \frac{(15+26)\mathbf{i} + (18+8)\mathbf{j}}{5} = \frac{41\mathbf{i}+26\mathbf{j}}{5}. \] Hence \( \vec{K} = 8.2\mathbf{i} + 5.2\mathbf{j} \).

(b) Resolve each leg into east (x) and north (y) components using bearing \( \theta \): east \(= d\sin\theta\), north \(= d\cos\theta\).

\( \vec{AB} = (8\sin 60^\circ,\ 8\cos 60^\circ) = (6.928,\ 4.000) \).
\( \vec{BC} = (12\sin 130^\circ,\ 12\cos 130^\circ) = (9.193,\ -7.713) \).

\( \vec{AC} = \vec{AB}+\vec{BC} = (16.121,\ -3.713) \).

(i) \( |AC| = \sqrt{16.121^2 + (-3.713)^2} = \sqrt{259.9+13.79} = \sqrt{273.7} = 16.5\text{ km} \) (3 s.f.).

(ii) The vector points east and south, so the bearing lies between \(90^\circ\) and \(180^\circ\): \[ \text{bearing} = 90^\circ + \tan^{-1}\!\left(\frac{3.713}{16.121}\right) = 90^\circ + 12.97^\circ \approx 103^\circ. \] The bearing of C from A is \(103^\circ\) (nearest degree).